1. Two dice are rolled. Let A, B, C be the events of getting a sum of 2, a sum of 3 and a sum of 4 respectively. Then, show that (i) A is a simple event (ii) B and C are compound events(iii) A and B are mutually exclusive

(ii) Since both B and C contain more than one sample point, each one of them is a compound event.

(iii) Since A ∩ B = ∅, A and B are mutually exclusive.

2. Two dice are rolled. A is the event that the sum of the numbers shown on the two dice is 5, and B is the event that at least one of the dice shows up a 3. Are the two events (i) mutually exclusive, (ii) exhaustive? Give arguments in support of your answer.

More examples related to the questions on the probabilities for throwing two dice.

3. Two dice are thrown simultaneously. Find the probability of: (i) getting six as a product(ii) getting sum ≤ 3(iii) getting sum ≤ 10(iv) getting a doublet(v) getting a sum of 8(vi) getting sum divisible by 5(vii) getting sum of atleast 11(viii) getting a multiple of 3 as the sum(ix) getting a total of atleast 10(x) getting an even number as the sum(xi) getting a prime number as the sum(xii) getting a doublet of even numbers(xiii) getting a multiple of 2 on one die and a multiple of 3 on the other die

Solution:

Two different dice are thrown simultaneously being number 1, 2, 3, 4, 5 and 6 on their faces. We know that in a single thrown of two different dice, the total number of possible outcomes is (6 × 6) = 36.

(i) getting six as a product:

Let E1 = event of getting six as a product. The number whose product is six will be E1 = [(1, 6), (2, 3), (3, 2), (6, 1)] = 4

Number of favorable outcomes
P(E11) =
Total number of possible outcome

= 15/36
= 5/12

(xii) getting a
doublet of even numbers:

Let E12 = event of getting a doublet of even numbers. The events of a doublet of even numbers will be E12 = [(2, 2), (4, 4), (6, 6)] = 3

Therefore, probability of
getting ‘a doublet of even numbers’

Number of favorable outcomes
P(E12) =
Total number of possible outcome

= 3/36
= 1/12

(xiii) getting a
multiple of 2 on one die and a multiple of 3 on the other die:

Let E13 = event of getting a multiple of 2 on one die and a multiple of 3 on the other die. The events of a multiple of 2 on one die and a multiple of 3 on the other die will be E13 = [(2, 3), (2, 6), (3, 2), (3, 4), (3, 6), (4, 3), (4, 6), (6, 2), (6, 3), (6, 4), (6, 6)] = 11

Therefore, probability of
getting ‘a multiple of 2 on one die and a multiple of 3 on the other die’

Number of favorable outcomes
P(E13) =
Total number of possible outcome

= 11/36

4. Two
dice are thrown. Find (i) the odds in favour of getting the sum 5, and (ii) the
odds against getting the sum 6.

Solution:

We know that in a single thrown of two die, the total number
of possible outcomes is (6 × 6) = 36.